Find the limits: (i) lim⁡x→1[x3−x2+1]\lim _{x \rightarrow 1}\left[x^{3}-x^{2}+1\right] \quad (ii) lim⁡x→3[x(x+1)]\lim _{x \rightarrow 3}[x(x+1)] (iii) lim⁡x→−1[1+x+x2+…+x10]\lim _{x \rightarrow-1}\left[1+x+x^{2}+\ldots+x^{10}\right].

Answer: (i) 1, (ii) 12, (iii) 1

Step-by-step solution

Step 1: Evaluate the first limit

For a polynomial function, the limit as xx approaches a certain value can be found by directly substituting that value into the function. In this case, we substitute x=1x=1 into the expression x3−x2+1x^3 - x^2 + 1.

Step 2: Calculate the first limit

After substituting x=1x=1, we perform the arithmetic operations: 13=11^3 = 1, 12=11^2 = 1. So the expression becomes 1−1+11 - 1 + 1, which simplifies to 11.

Step 3: Evaluate the second limit

Similarly, for the second limit, we have a polynomial function x(x+1)x(x+1). We can find the limit as xx approaches 33 by directly substituting x=3x=3 into the expression.

Step 4: Calculate the second limit

After substituting x=3x=3, the expression becomes 3(3+1)3(3+1), which simplifies to 3(4)3(4), resulting in 1212.

Step 5: Evaluate the third limit

The third limit involves a sum of powers of xx, which is also a polynomial. We substitute x=−1x=-1 into each term of the sum.

Step 6: Calculate the third limit

When x=−1x=-1, the terms alternate between 11 and −1-1. Specifically, 11, (−1)1=−1(-1)^1 = -1, (−1)2=1(-1)^2 = 1, and so on. Since there are 1111 terms in total (from x0x^0 to x10x^{10}), and pairs of (1−1)(1-1) cancel out, the last term (−1)10(-1)^{10} is 11, leaving a final sum of 11.

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